# Is it possible to evaluate $\int_0^1 x^n \, dx$ by contour integration?

It's been quite a time since I had the complex analysis course. The thing is now I don't know the answer to the following simple question:

Is it possible to find

$$\int_0^1 x^n \, dx$$

using the methods of contour integration? I've refreshed my knowledge with wikipedia, but I've no idea how to make integrals with no obvious singularities. Though there is a singularity at $\infty$, but how to connect it with $[0,1]$?

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Well you need to define a closed path, of which $[0,1]$ is a part. For example, you could take the parametrization $$z(t)=\left\{\matrix{it&&t<0\\t&&0\le t\le1\\1+i(t-1)&\quad&t>1}\right.$$ which is closed since the path meets at the point at infinity. Then if you can integrate $z^n$ on the first and third parts, you can use the fact that the sum over all three parts vanishes. –  bgins Jun 10 '12 at 8:07
I think he wants an answer involving the residue theorem for a slick computation. –  Potato Jun 10 '12 at 8:26
@bgins I don't think I would be able to integrate on those infinite contour parts. And yes, Potato is right. –  Yrogirg Jun 10 '12 at 8:31
@Yrogirg I strongly doubt a nice application of the residue theorem is possible. The usual classes of integrals that the residue theorem is applied to can be found in Churchill or Ahlfors. –  Potato Jun 10 '12 at 8:34
A possible candidate is $$- \int_{C_r} \dfrac{\log(1-z)}{z^{n+2}} dz.$$ But I have not yet found how to relate it to $\int_0^1 x^n dx$. –  user17762 Jun 10 '12 at 9:29

Let's consider the contour from $0$ to $1$ ($z=x$) followed by a rotation of angle $\angle \frac{2\pi}n$ ($z=e^{i\phi}$) and then back to $0$ ($z= e^{i\frac {2\pi}n} x$) :

$$\int_0^1 x^n\,dx+\int_0^{\frac {2\pi}n} e^{ni\phi}ie^{i\phi}\,d\phi +\int_1^0 \left(e^{i\frac {2\pi}n}x\right)^n e^{i\frac {2\pi}n}\,dx=0$$

(the integral is $0$ of course for $n\ne -1$)

since $\displaystyle \int_0^{\frac {2\pi}n} e^{ni\phi}ie^{i\phi}\,d\phi=\left[\frac{e^{(n+1)i\phi}}{n+1}\right]^{\frac {2\pi}n}_{\phi=0}$ we get : $$\left(1-e^{i\frac {2\pi}n}\right)\int_0^1 x^n\,dx=\frac {1-e^{i\frac {2\pi}n}}{n+1}$$

so that for $n\ne 1$ and $n\ne -1$ at least : $\ \displaystyle \int_0^1 x^n\,dx=\frac 1{n+1}$

For $n=1$ we may choose another maximal angle (for example $\pi$ or $\frac {\pi}2$).
Looking back at this there is some feeling of cheating since we replaced a power integral over $x^n$ by the nearly equivalent integral $\int e^{(n+1)i\phi}\,d\phi$ (at least the increment of $n$ was done!).

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This is rather a teaser than a proper answer. Well formally it is an answer since it employs contour integration, but not quite in a way I was hoping to see it.

While I was asking about $x^n$, actually I was heading for polynomials in general. So let's pick $P(x)$ to be a polynomial and consider

$$\int_a^b P(x) \, dx$$

Following @Marvis 's idea antiderivative of $P$ is computed as

$$\int P(x) \, dx = - \int_{C_r} x P\left( \frac{x}{z} \right) \frac{\log{(1-z)}}{z^2} \, dz$$

Then it's just a fundamental theorem of calculus to find the initial integral. Note, that $x$ above is just a parameter, not the real part of $z$.

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